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Chapter 6

Chi-Square Test for Categorical Variable

Shaoqi Rao, PhD

2009.11.9

Slides adapted from Dr. Zhang Jinxin’s

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6.1 Basic logic of 6.1 Basic logic of 22 test testGiven a set of observed frequency distribution

A1, A2, A3 …

to test whether the data follow certain theory.If the theory is true, then we will have a set

of theoretical frequency distribution:

T1, T2, T3 …

Comparing A1, A2, A3 … and T1, T2, T3 …

If they are quite different, then the theory might not be true;

Otherwise, the theory is acceptable.

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6.1.16.1.1 Chi-square distribution Chi-square distribution

~ 2 distribution

—— Agreement between observed and expected frequencies

k

i i

iiP e

ef

1

22 )(

DF=k-1-# parameters estimating fi

For a contingency table,

DF=(# rows-1)(# columns-1）

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22 distributiondistribution

0 2 4 6 8 100.0

0.1

0.2

0.3

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6.1.2 χ2Test for Goodness of Fit (Large Sample)

Table1 Frequency distribution and goodness of fit based on 136 measurements to the phantom( 体模 )

intervals A Φ(X1) Φ(X2) P(X) T=n* P(X) (A-T)2/T

1.228- 2 0.00069 0.00466 0.00397 0.5405 3.94143

1.234- 2 0.00466 0.02275 0.01809 2.4601 0.08605

1.240- 7 0.02275 0.08076 0.05801 7.8889 0.10016

1.246- 17 0.08076 0.21186 0.13110 17.8294 0.03859

1.252- 25 0.21186 0.42074 0.20888 28.4083 0.40892

1.258- 37 0.42074 0.65542 0.23468 31.9167 0.80961

1.264- 25 0.65542 0.84134 0.18592 25.2855 0.00322

1.270- 16 0.84134 0.94520 0.10386 14.1244 0.24906

1.276- 4 0.94520 0.98610 0.04090 5.5618 0.43858

1.282- 1 0.98610 0.99744 0.01135 1.5434 0.19130

合 计 - - - - - 6.26692

00.201.0

26.1240.1

Z 40.1

01.0

26.1246.1

Z

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1. Setting up hypothesesH0 ： the population follows N(1.26,0.012)

H1 ： the population doesn’t follow N(1.26,0.012) α=0.05

2. Calculation of the statistic ：

3. P-value ： ν=k-1-2=10-1-2=7

4. Conclusion ： With significance level α=0.05, H0 is not rejected. The measurement follows the normal distribution.

27.6

22

T

TA

5.0

07.14,35.62

7,5.02

27,05.0

27,5.0

P

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6.2 Comparison between Two Independent

Sample Proportions

In chapter 4 the Z test can only be used

for comparing with a given 0 (one sample)

or comparing 1 with 2 (two samples).

If we need to compare more than two

samples, Chi-square test is widely used.

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Example 6.1Example 6.1 In a clinical survey, 215 patients with pulmona

ry heart disease ( 肺心病 ) in a hospital were collected , of which 164 patients have taken digitalis ( 洋地黄 ) and 51 patients haven’t taken it. Each of them received an ECG examination. The results are listed in Table 6.2.

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Table 6.2 Data of patients of pulmonary heart disease with arrhythmia（心律失常）

ECG

Arrhythmia Normal Total Arrhythmia rate (%)

With digitalis 81 83 164 49.39

Without digitalis 19 32 51 37.25

Total 100 115 215 46.51

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Table 6.2 Data of patients of pulmonary heart disease with arrhythmia（心律失常）

ECG

Arrhythmia Normal Total Arrhythmia rate (%)

With digitalis 81(76.28) 83(87.72) 164 49.39

Without digitalis 19(23.72) 32(27.28) 51 37.25

Total 100 115 215 46.51

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2P ＝

11510051164

215)19833281( 2

=2.3028

2

1

2

1

22 )(

i j ij

ijijP e

ef

3028.228.27

)28.2732(

72.23

)72.2319(

72.87

)72.8783(

28.76

)28.7681( 22222

p

2121

2211222112 )(

ccrrp nnnn

nffff

ν ＝ 1

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22 test and test and ZZ test test

According to (4.25)

5175.1)51/1164/1)(215/115)(215/100(

51/19164/81

Z

3028.22 Z）（ 25.4

11)1(

2100

21

nnPP

PPz

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Correction for continuityCorrection for continuity

When n≥40, if there happens 1≤eij<5,

2

1

2

1

2

2)5.0(

i j ij

ijij

P e

ef

2121

2211222112 )2/(

ccrrP nnnn

nnffff

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Fisher’s exact testFisher’s exact test

When n<40, or eij<1, with SPSS, 2 test is not proper then. An exact P value will be obtained for us to give conclusion.

This can be easily fulfilled in SPSS.

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Example 6.9Example 6.9

Table 6.14 The results of treatment to embolic angitis（栓塞性脉管炎）patients Groups Recovery No recovery Total

New treatment 6(a) 1(b) 7(nr1) Control 1(c) 4(d) 5(nr2)

Total 7(nc1) 5(nc2) 12(n)

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Statistical descriptionStatistical description

group * result Crosstabulation

6 1 7

4.1 2.9 7.0

85.7% 14.3% 100.0%

1 4 5

2.9 2.1 5.0

20.0% 80.0% 100.0%

7 5 12

7.0 5.0 12.0

58.3% 41.7% 100.0%

Count

Expected Count

% within group

Count

Expected Count

% within group

Count

Expected Count

% within group

new treatment

control

group

Total

recovery no recovery

result

Total

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Chi-Square Tests

5.182b 1 .023

2.831 1 .092

5.555 1 .018

.072 .045

4.750 1 .029

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Pearson Chi-Square

Continuity Correctiona

Likelihood Ratio

Fisher's Exact Test

Linear-by-LinearAssociation

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)Exact Sig.(2-sided)

Exact Sig.(1-sided)

Computed only for a 2x2 tablea.

4 cells (100.0%) have expected count less than 5. The minimum expected count is2.08.

b.

Statistical inferenceStatistical inference

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6.3 The 6.3 The 22 Tests for Binary Tests for Binary Variable under a Paired DesignVariable under a Paired Design

Example 6.2 There are 260 serum ( 血清 ) samples. Each sample is divided into two and tested by two different methods of immunological test of rheumatoid factor( 类风湿因子 ) respectively. The results are listed in Table 6.4. Now the question is that results of two methods are independent or not.

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test for independence between test for independence between two binary variablestwo binary variables

Table 6.4 The results of two immunological tests B

A ＋ －

Total

＋ 172 8 180 － 12 68 80

Total 184 76 260

2121

2211222112 )2/(

ccrrP nnnn

nnffff 2 2 =173.74=173.74

Example 6.2Example 6.2

12/80=15%12/80=15%172/180=95%172/180=95%

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6.3.2 Comparison between 6.3.2 Comparison between two sample proportionstwo sample proportions

McNemar testMcNemar test

2112

22112 )(

ff

ff

2 2 ==

21

H0: 1=2, H1: 1≠2, α=0.05When H0 is true,

For large sample (b+c>40)

If the 2 > 2 , then reject H0

221

cbTT

cb

cbcb

cbc

cb

cbb

222

2 )(

2

)2

(

2

)2

(

0.05

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The Probability ExpressionsThe Probability Expressions

Trt A Trt B Total

+ -

+ 11 (a) 12 (b) r1

- 21 (c) 22 (d) r2

Total c1 c2 1.0

H0: c1= r1 H1: c1 r1

Since c1= 11+ 21, r1= 11+ 12,

This test becomes: H0: 12= 21, H1: 12 21

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Correction to McNemar testCorrection to McNemar test((ff1212++ff2121<40)<40)

2112

22112 )1(

ff

ff

2 2 == 112

128

)1128( 2

2 2 = =0.45= =0.45

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6.4 The 6.4 The 22 Test for R×C Test for R×C Contingency TableContingency Table

Table 6.6 Blood types of patient suffering from different diseases Blood type Total

Disease status A B O

Digestive ulcer 679 134 983 1796 Stomach cancer 416 84 383 883

Control 2625 570 2892 6087 Total 3720 788 4258 8766

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The statistic for hypothesis testThe statistic for hypothesis test

i j cjri

ij

nn

fn )1(

2

2 2 ==

543.40142586087

2892

37201796

6798766

222

P

4)1()1( CR

205.0 =9.488=9.488

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6.4.2 Multiple comparison6.4.2 Multiple comparison for R×C Table for R×C Table

group + －

I

II

III

IV

V

…

…

…

…

…

…

…

…

…

…

VI … …controlcontrol

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6.4.3 6.4.3 Measurement of Measurement of association for R×C tableassociation for R×C table

Table 6.11 Blood type of 1043 patients MN system ABO

system M N MN Total

O 85 100 150 335

A 56 78 120 254

B 98 132 170 400 AB 23 25 6 54

Total 262 335 446 1043

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Pearson contingency coefficientPearson contingency coefficient

2

2

P

PP nr

156.0925.251043

925.25

Pr

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Pre-requisite for 2 test

By experience, The theoretical frequencies should be grea

ter than 5 in more than 4/5 cells; The theoretical frequency in any cell shoul

d be greater than 1.

Otherwise, we need to use Fisher exact test.

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